We need to clarify what the equation means. If you're integrating $\mathrm{dx}$ then $x$ is a variable so it is not allowed to appear in the integration limits. Nor is it then allowed to appear on the LHS of the equation. I think what you mean is this:

Find all functions $y(x)$ which satisfy (for every fixed $x$):

$x={\int }_1^x(1+y\text{'}(t))\mathrm{dt}$

where $y\text{'}(t)$ represents $\mathrm{dy}/\mathrm{dx}$ at $x=t$.

If this is what you mean, then the thing to do is differentiate each side with respect to $x$. (We're allowed to do this as the equation is an identity in $x$.) The RHS comes to $1+y\text{'}(x)$ by the fundamental theorem of calculus. We get:

$1=1+y\text{'}(x)$

Same idea. You can differentiate both sides, and using the fundamental theorem of calculus:

$1=(1+y\text{'}(x{)}^2{)}^{1/2}$

So $y\text{'}(x{)}^2=0$ and again $y=\mathrm{const}$ is the only possibility.

Are you happy with the fundamental theorem of calculus step? The fact we need to use is that if:

$z(x)={\int }_0^{x}f(t)\mathrm{dt}$

If you let $p(x)$ be the integral of $f(t)$ from 100 to $x$ and let $q(x)$ be the integral of $f(t)$ from 0 to $x$ then:

$q(x)=p(x)+{\int }_0^{100}f(t)\mathrm{dt}$

The last term on the RHS is a constant. So differentiating we get: